Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{z^2 - 12z + 35}{z + 3} \div \dfrac{z^2 - 7z}{-7z - 21} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{z^2 - 12z + 35}{z + 3} \times \dfrac{-7z - 21}{z^2 - 7z} $ First factor the quadratic. $n = \dfrac{(z - 7)(z - 5)}{z + 3} \times \dfrac{-7z - 21}{z^2 - 7z} $ Then factor out any other terms. $n = \dfrac{(z - 7)(z - 5)}{z + 3} \times \dfrac{-7(z + 3)}{z(z - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (z - 7)(z - 5) \times -7(z + 3) } { (z + 3) \times z(z - 7) } $ $n = \dfrac{ -7(z - 7)(z - 5)(z + 3)}{ z(z + 3)(z - 7)} $ Notice that $(z + 3)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ -7\cancel{(z - 7)}(z - 5)(z + 3)}{ z(z + 3)\cancel{(z - 7)}} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $n = \dfrac{ -7\cancel{(z - 7)}(z - 5)\cancel{(z + 3)}}{ z\cancel{(z + 3)}\cancel{(z - 7)}} $ We are dividing by $z + 3$ , so $z + 3 \neq 0$ Therefore, $z \neq -3$ $n = \dfrac{-7(z - 5)}{z} ; \space z \neq 7 ; \space z \neq -3 $